Help with Prince Rupert's Problem Prince Rupert's problem asks how large an m dimensional cube can be inscribed in an n dimensional unit cube.
For $m=1$ and $m=2, n=3$, this is pretty easy, where $f(m,n)$ is the edge of the cube.
$$
f(1,n) = \sqrt n
$$
$$
f(2,3)=\frac 3 4 \sqrt 2
$$
From there it's supposed to be straightforward to show:
$$
f(m+1,n) < f(m,n) < f(m,n+1)
$$
While this is easy for the case where m=1 I can't get any further than that.
 A: Consider the (enlarged) cube $A$ in $\mathbb R^n$ spanned by the $2^n$ vertices $v$ of the form $|v^{(i)}|=1$, $1\le i\le n$.
In it consider, for $1\le k\le n$,  an $(n-1)$ dimensional cube $B_k$  such that all $2^{n-1}$ vertices $w$ of $B_k$ have the properties


*

*$|w^{(i)}|<1$ for $1\le i\le k$

*$|w^{(i)}|\le 1$ for $k< i\le n$


The cube $B_1$ can be obtained by projecting $A$ into the orthogonal complement of the first standard base vector so that we even have $w^{(0)}=0$.
Assume we have $B_k$ for some $k<n$.
Also assume that we have an orthogonal map $\Psi_k$ such that $B_k=\Psi_k B_1$ and that $\Psi_k e_i=e_i$ for $i> k$. (Thus we can take $\Psi_1=1$).
For $t\in\mathbb R$, we can consider the rotation $\Phi_t$ that maps $e_1\mapsto \cos( t) e_1+\sin(t)e_{k+1}$, $e_{k+1}\mapsto\cos(t)e_{k+1}-\sin(t)e_1$ and $e_i\mapsto e_i$ for $i\notin\{1,k+1\}$.
Let $w$ be a vertex of $B_1$ and let $u=\Psi_k\Phi_t w$.
Then $u$ is a vertex of $B_k$ if $t=0$.
By continuity of $t\mapsto\Phi_t$, we have $|u^{(i)}|<1$ for $1\le i\le k$, provided $|t|$ is small enough.
Also, $u^{(i)}=w^{(i)}$ if $i>k+1$.
We have $u=\Psi_k w + \Psi_k(\Phi_1-1)w\approx \Psi_k w -tw^{(k+1)} \Psi_k e_{k+1}$, hence
$u^{(k+1)}\approx (1-t)w^{(k+1)} $, that is for small positive $t$, we will have $|u^{(k+1)}|\approx1-t$ so that $B_{k+1}:= \Psi_{k+1}B_1$ with $\Psi_{k+1}:=\Psi_k\Phi_t$ has the desired properties.
Then $B_n$ is an $(n-1)$-dimensional hypercube that fits strictly inside the $n$-dimensional hypercube $A$. In other words, $B_n$ can be stretched by an amount $q>1$ and still fits.
This shows that 
$$ f(n,n+1)>1\quad\text{for all }n\in\mathbb N.$$
Since trivially 
$$f(m,n)\ge f(m,k)f(k,n)\quad\text{if }m\le k\le n,$$
we arrive at 
$$ f(m+1,n)<f(m,m+1)f(m+1,n)\le f(m,n)$$
at least if $m+1\le n$ (i.e. if $f(m+1,n)> 0$).
We also find 
$$f(m,n)< f(m,n)f(n,n+1)\le f(m,n+1)$$
if $m\le n$ (so that $f(m,n)>0$). Since trivially $0=f(m+1,n)<f(m,n)=1$ if $m=n$, we finally obtain
$$ f(m+1,n)<f(m,n)<f(m,n+1)\quad\text{if }1\le m\le n.$$
